A town has a population of 1 million. The average electric power...

Total electric power required by the town:

$P_e = 10^6 \times 300 \text{ W} = 3 \times 10^8 \text{ W}$

Since the efficiency of converting thermal power to electric power is $25\%$, the total thermal power required is:

$P_{th} = \dfrac{P_e}{0.25} = \dfrac{3 \times 10^8}{0.25} = 1.2 \times 10^9 \text{ W} = 1.2 \times 10^9 \text{ J/s}$

Energy released per fission of ^{235}\text{U}$is:<br><br>$E = 200 \text{ MeV} = 200 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 3.2 \times 10^{-11} \text{ J}$<br><br>The number of fissions occurring per second is:<br><br>$\dfrac{dn}{dt} = \dfrac{P_{th}}{E} = \dfrac{1.2 \times 10^9}{3.2 \times 10^{-11}} = 3.75 \times 10^{19} \text{ s}^{-1}$<br><br>The number of fissions per day is:<br><br>$N = 3.75 \times 10^{19} \times (24 \times 60 \times 60) = 3.75 \times 10^{19} \times 86400 = 3.24 \times 10^{24}$<br><br>The number of moles of ^{235}\text{U}$ consumed per day is:

$n = \dfrac{N}{N_A} = \dfrac{3.24 \times 10^{24}}{6.022 \times 10^{23}} \approx 5.38 \text{ mol}$

The mass of ^{235}\text{U}$consumed per day is:<br><br>$m = n \times 235 \text{ g} = 5.38 \times 235 \text{ g} \approx 1264.3 \text{ g} = 1.264 \text{ kg}